Prove that 0.5(y2−x2)≤y(y−x) and 0.5(y2−x2)≥x(y−x)

Question

Let 0≤a<b and consider the function h(x)=x on [a,b]. Prove that for any 0≤x≤y we have 0.5(y2−x2)≤y(y−x) and 0.5(y2−x2)≥x(y−x)

Answer 1

1/2(y-x)^2 >= 0 for any y and x.

Then, x(y-x) = 1/2(x+x)(y-x) <= 1/2(x+x)(y-x) + 1/2(y-x)^2 = 1/2(x+x + y-x)(y-x) = 1/2(x+y)(y-x) = 1/2(y^2 - x^2) = 1/2(x+y)(y-x) <= 1/2(x+y)(y-x) + 1/2(y-x)^2 = 1/2(x+y+y-x)(y-x) = 1/2(2y)(y-x) = y(y-x)

x(y-x) <= 1/2(y^2 - x^2) <= y(y-x)

No conditions on x and y are needed in this problem.

Answer 2

0.5(y + x)(y - x) ≤ y(y - x)

0.5(y + x) ≤ y division by y - x > 0

0.5y + 0,5x ≤ y

0.5x ≤ 0.5y

x ≤ y

QED

Do the second part in the same manner.